Calculate the wavelengths of the first five members of the Lyman series of spectral lines, providing the result in units Angstrom with precision one digit after the decimal point.

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Respuesta :

Answer:

Explanation:

The formula for hydrogen atomic  spectrum is as follows

energy of photon due to transition from higher orbit n₂ to n₁

[tex]E=13.6 (\frac{1}{n_1^2 } - \frac{1}{n_2^2})eV[/tex]

For layman series n₁ = 1 and n₂ = 2 , 3 , 4 ,   ...   etc

energy of first line

[tex]E_1=13.6 (\frac{1}{1^2 } - \frac{1}{2 ^2})[/tex]

10.2 eV

wavelength of photon = 12375 / 10.2 = 1213.2 A

energy of 2 nd line

[tex]E_2=13.6 (\frac{1}{1^2 } - \frac{1}{3 ^2})[/tex]

= 12.08 eV

wavelength of photon = 12375 / 12.08 = 1024.4 A

energy of third line

[tex]E_3=13.6 (\frac{1}{1^2 } - \frac{1}{4 ^2})[/tex]

12.75 e V

wavelength of photon = 12375 / 12.75 = 970.6 A

energy of fourth line

[tex]E_4=13.6 (\frac{1}{1^2 } - \frac{1}{5 ^2})[/tex]

= 13.056 eV

wavelength of photon = 12375 / 13.05 = 948.3 A

energy of fifth line

[tex]E_5=13.6 (\frac{1}{1^2 } - \frac{1}{6 ^2})[/tex]

13.22 eV

wavelength of photon = 12375 / 13.22 = 936.1 A